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Question Number 152151 by john_santu last updated on 26/Aug/21

	Answered by Rasheed.Sindhi last updated on 26/Aug/21

	$f (x) = \frac{a x + b}{c x + d}$ $∙ f (f (x)) = \frac{a (\frac{a x + b}{c x + d}) + b}{c (\frac{a x + b}{c x + d}) + d} = x$ $= \frac{a (a x + b) + b (c x + d)}{c (a x + b) + d (c x + d)} = x$ $x = 0 : \frac{a b + b d}{b c + d^{2}} = 0$ $b (a + d) = 0; b c + d^{2} \neq 0$ $b = 0 ∣ a = - d$ $∙ f (19) = \frac{19 a + b}{19 c + d} = 19; \frac{97 a + b}{97 c + d} = 97$ $19 a + b = 19^{2} c + 19 d) \times 97$ $97 a + b = 97^{2} c + 97 d) \times 19$ $78 b = (19^{2} . 97 - 97^{2} . 19) c$ $b = 0 \Rightarrow c = 0$ $b = c = 0, a = - d$ $19 a + b = 19^{2} c + 19 d \Rightarrow 19 (- d) = 19 d$ $\Rightarrow d = 0 \Rightarrow a = - d \Rightarrow a = 0$ $a = b = c = d = 0 ? ? ? ? f (59) = ?$
	Commented by john_santu last updated on 26/Aug/21

	$no sir .$
	Commented by prakash jain last updated on 27/Aug/21

	$If we put b = d = 0$ $\frac{19 a}{d} = 19 \Rightarrow a = d$ $f (x) = x$ $b = 0 or a = - d$ $so when b = 0 a need not be - d$
	Commented by Rasheed.Sindhi last updated on 27/Aug/21

	$THANKS SIR!$
	Answered by prakash jain last updated on 26/Aug/21

	$f (x) = x \Rightarrow f (f (x)) = x$ $f (59) = 59$ $a = 1$ $b = c = 0$ $d = 1$
	Commented by Rasheed.Sindhi last updated on 26/Aug/21

	$S i r, i f b = c = d = 0$ $f (x) = \frac{a x + b}{c x + d} = \frac{1 x + 0}{0 x + 0} = \frac{x}{0} = ?$
	Commented by prakash jain last updated on 26/Aug/21

	$Thanks .$ $Correctrd$ $b = c = 0$ $d = 1$ $so that f (x) = x$
	Commented by john_santu last updated on 27/Aug/21

	$wrong sir$
	Commented by prakash jain last updated on 06/Sep/21

	$You are right . I did not read question$ $properly . Question specifically said$ $a, b, c, d a r e n o t z e r o .$
	Answered by john_santu last updated on 27/Aug/21
	$For all x ,f(f(x))=x ,i.e ((a(((ax+b)/(cx+d)))+b)/(c(((ax+b)/(cx+d)))+d)) = x, i.e (((a^2 +bc)x+b(a+d))/(c(a+d)x+bc+d^2 )) = x , i.e c(a+d)x^2 +(d^2 −a^2 )x−b(a+d)=0, which implies c(a+d)=0 , since c ≠0 we must have a=−d . The conditions f(19)=19 and f(97)=97 lead to the equations { ((19^2 c=2.19a+b)),((97^2 c=2.97a+b)) :} Hence (97^2 −19^2 )c = 2(97−19)a it follows that a=18c ,which in turn leads to b=−1843c therefore f(x)=((58x−1843)/(x−58)) so f(59)=((58×59−1843)/(59−58)) =1579$
	$For all x, f (f (x)) = x, i . e$ $\frac{a (\frac{ax + b}{cx + d}) + b}{c (\frac{ax + b}{cx + d}) + d} = x, i . e$ $\frac{(a^{2} + bc) x + b (a + d)}{c (a + d) x + bc + d^{2}} = x, i . e$ $c (a + d) x^{2} + (d^{2} - a^{2}) x - b (a + d) = 0,$ $which implies c (a + d) = 0, since c \neq 0$ $we must have a = - d .$ $The conditions f (19) = 19 and f (97) = 97$ $lead to the equations$ ${\begin{cases} 19^{2} c = 2 . 19 a + b \\ 97^{2} c = 2 . 97 a + b \end{cases}$ $Hence (97^{2} - 19^{2}) c = 2 (97 - 19) a$ $it follows that a = 18 c, which in$ $turn leads to b = - 1843 c$ $therefore f (x) = \frac{58 x - 1843}{x - 58}$ $so f (59) = \frac{58 \times 59 - 1843}{59 - 58} = 1579$
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